Basic Limit Exercises
edit
Evaluate the following limits or state that the limit does not exist.
8.
lim
x
→
−
1
−
1
−
x
2
{\displaystyle \lim _{x\to -1^{-}}{\sqrt {1-x^{2}}}}
The limit does not exist.
The limit does not exist.
Solutions
Evaluate the following limits or state that the limit does not exist.
12.
lim
x
→
4
1
x
−
4
{\displaystyle \lim _{x\to 4}{\frac {1}{x-4}}}
The limit does not exist.
The limit does not exist.
13.
lim
x
→
2
1
x
−
2
{\displaystyle \lim _{x\to 2}{\frac {1}{x-2}}}
The limit does not exist.
The limit does not exist.
20.
lim
x
→
0
|
x
|
x
{\displaystyle \lim _{x\to 0}{\frac {|x|}{x}}}
The limit does not exist.
The limit does not exist.
22.
lim
x
→
3
x
2
+
16
x
−
3
{\displaystyle \lim _{x\to 3}{\frac {\sqrt {x^{2}+16}}{x-3}}}
The limit does not exist.
The limit does not exist.
25.
lim
x
→
3
x
+
3
x
2
−
9
{\displaystyle \lim _{x\to 3}{\frac {x+3}{x^{2}-9}}}
The limit does not exist.
The limit does not exist.
31.
lim
x
→
1
5
x
x
2
+
2
x
−
3
{\displaystyle \lim _{x\to 1}{\frac {5x}{x^{2}+2x-3}}}
The limit does not exist.
The limit does not exist.
32.
lim
x
→
−
1
ln
(
x
2
−
1
)
{\displaystyle \lim _{x\to -1}\ln({\sqrt {x^{2}-1}})}
The limit does not exist.
The limit does not exist.
33.
lim
x
→
1
arcsin
(
x
)
{\displaystyle \lim _{x\to 1}\arcsin(x)}
The limit does not exist.
The limit does not exist.
Solutions
Evaluate the following limits or state that the limit does not exist.
Solutions
Limits of Piecewise Functions
edit
Evaluate the following limits or state that the limit does not exist.
48. Consider the function
f
(
x
)
=
{
(
x
−
2
)
2
if
x
<
2
x
−
3
if
x
≥
2
{\displaystyle f(x)={\begin{cases}(x-2)^{2}&{\text{if }}x<2\\x-3&{\text{if }}x\geq 2\end{cases}}}
c.
lim
x
→
2
f
(
x
)
{\displaystyle \lim _{x\to 2}f(x)}
49. Consider the function
g
(
x
)
=
{
−
2
x
+
1
if
x
≤
0
x
+
1
if
0
<
x
<
4
x
2
+
2
if
x
≥
4
{\displaystyle g(x)={\begin{cases}-2x+1&{\text{if }}x\leq 0\\x+1&{\text{if }}0<x<4\\x^{2}+2&{\text{if }}x\geq 4\end{cases}}}
50. Consider the function
h
(
x
)
=
{
2
x
−
3
if
x
<
2
8
if
x
=
2
−
x
+
3
if
x
>
2
{\displaystyle h(x)={\begin{cases}2x-3&{\text{if }}x<2\\8&{\text{if }}x=2\\-x+3&{\text{if }}x>2\end{cases}}}
Solutions
51. Use the intermediate value theorem to show that there exists a value
f
(
x
)
=
3
{\displaystyle f(x)=3}
for
f
(
x
)
=
x
+
1
x
{\displaystyle f(x)={\sqrt {x}}+{\frac {1}{\sqrt {x}}}}
from
1
≤
x
≤
9
{\displaystyle 1\leq x\leq 9}
. If you cannot use the intermediate value theorem to show this, explain why.
Notice
f
(
x
)
{\displaystyle f(x)}
x
∈
[
1
,
9
]
{\displaystyle x\in \left[1,9\right]}
y
∈
[
2
,
10
3
]
{\displaystyle y\in \left[2,{\frac {10}{3}}\right]}
c
∈
(
1
,
9
)
{\displaystyle c\in (1,9)}
f
(
c
)
=
3
{\displaystyle f(c)=3}
◼
{\displaystyle \blacksquare }
Notice
f
(
x
)
{\displaystyle f(x)}
x
∈
[
1
,
9
]
{\displaystyle x\in \left[1,9\right]}
y
∈
[
2
,
10
3
]
{\displaystyle y\in \left[2,{\frac {10}{3}}\right]}
c
∈
(
1
,
9
)
{\displaystyle c\in (1,9)}
f
(
c
)
=
3
{\displaystyle f(c)=3}
◼
{\displaystyle \blacksquare }
52. Use the intermediate value theorem to show that there exists an
x
=
μ
{\displaystyle x=\mu }
so that
f
(
μ
)
=
π
4
{\displaystyle f(\mu )={\frac {\pi }{4}}}
for
f
(
x
)
=
sin
(
x
)
−
cos
(
x
)
ln
x
{\displaystyle f(x)={\frac {\sin(x)-\cos(x)}{\ln {x}}}}
from
π
≤
x
≤
9
4
π
{\displaystyle \pi \leq x\leq {\frac {9}{4}}\pi }
. If you cannot use the intermediate value theorem to show this, explain why.
Notice
f
(
x
)
{\displaystyle f(x)}
x
∈
[
π
,
9
4
π
]
{\displaystyle x\in \left[\pi ,{\frac {9}{4}}\pi \right]}
f
(
π
)
=
1
ln
(
π
)
{\displaystyle f(\pi )={\frac {1}{\ln(\pi )}}}
f
(
9
4
π
)
=
0
{\displaystyle f\left({\frac {9}{4}}\pi \right)=0}
It is known the following is true:
1
≥
π
4
≥
0
{\displaystyle 1\geq {\frac {\pi }{4}}\geq 0}
1
ln
(
π
)
≥
1
⇒
1
ln
(
π
)
≥
1
≥
π
4
≥
0
⇒
1
ln
(
π
)
≥
π
4
≥
0
{\displaystyle {\begin{aligned}{\frac {1}{\ln(\pi )}}\geq 1&\Rightarrow {\frac {1}{\ln(\pi )}}\geq 1\geq {\frac {\pi }{4}}\geq 0\\&\Rightarrow {\frac {1}{\ln(\pi )}}\geq {\frac {\pi }{4}}\geq 0\end{aligned}}}
f
(
x
)
{\displaystyle f(x)}
x
∈
[
π
,
9
4
π
]
{\displaystyle x\in \left[\pi ,{\frac {9}{4}}\pi \right]}
x
=
μ
{\displaystyle x=\mu }
f
(
9
4
π
)
≤
f
(
μ
)
≤
f
(
π
)
{\displaystyle f\left({\frac {9}{4}}\pi \right)\leq f(\mu )\leq f(\pi )}
μ
∈
[
π
,
9
4
π
]
{\displaystyle \mu \in \left[\pi ,{\frac {9}{4}}\pi \right]}
◼
{\displaystyle \blacksquare }
Notice
f
(
x
)
{\displaystyle f(x)}
x
∈
[
π
,
9
4
π
]
{\displaystyle x\in \left[\pi ,{\frac {9}{4}}\pi \right]}
f
(
π
)
=
1
ln
(
π
)
{\displaystyle f(\pi )={\frac {1}{\ln(\pi )}}}
f
(
9
4
π
)
=
0
{\displaystyle f\left({\frac {9}{4}}\pi \right)=0}
It is known the following is true:
1
≥
π
4
≥
0
{\displaystyle 1\geq {\frac {\pi }{4}}\geq 0}
1
ln
(
π
)
≥
1
⇒
1
ln
(
π
)
≥
1
≥
π
4
≥
0
⇒
1
ln
(
π
)
≥
π
4
≥
0
{\displaystyle {\begin{aligned}{\frac {1}{\ln(\pi )}}\geq 1&\Rightarrow {\frac {1}{\ln(\pi )}}\geq 1\geq {\frac {\pi }{4}}\geq 0\\&\Rightarrow {\frac {1}{\ln(\pi )}}\geq {\frac {\pi }{4}}\geq 0\end{aligned}}}
f
(
x
)
{\displaystyle f(x)}
x
∈
[
π
,
9
4
π
]
{\displaystyle x\in \left[\pi ,{\frac {9}{4}}\pi \right]}
x
=
μ
{\displaystyle x=\mu }
f
(
9
4
π
)
≤
f
(
μ
)
≤
f
(
π
)
{\displaystyle f\left({\frac {9}{4}}\pi \right)\leq f(\mu )\leq f(\pi )}
μ
∈
[
π
,
9
4
π
]
{\displaystyle \mu \in \left[\pi ,{\frac {9}{4}}\pi \right]}
◼
{\displaystyle \blacksquare }
Solutions
![{\displaystyle \lim _{x\to 2}{\Big [}4x^{2}-3x-1{\Big ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b9090ef0539e55c8c495638167af1902f6c0371e)
![{\displaystyle \lim _{x\to 5}{\Big [}x^{2}{\Big ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ddef2e7715fea4bcb0a1fdacbcc14b5049a50a46)
![{\displaystyle \lim _{x\to {\frac {\pi }{4}}}{\Big [}\cos ^{2}(x){\Big ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b2b628dde4a64c213ceea9a932bba29f50ebe270)
![{\displaystyle \lim _{x\to 1}{\Big [}5e^{x-1}-5{\Big ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1fbd3c44c22819b3d1cfe0b427d665f4742f96bc)
![{\displaystyle \lim _{x\to 7^{-}}{\Big [}|x^{2}+x|-x{\Big ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/107d70cb67026562a0c3ed9db454d47178947b2b)
![{\displaystyle \lim _{x\to 1}\left[x^{2}+5x-{\frac {1}{2-x}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00ff82956e4812c90be92af6b92fe043d09fec3c)
![{\displaystyle \lim _{x\to -\infty }{\Big [}3x^{2}-2x+1{\Big ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5a9cfb494972698a81cd5a92200191100d2baff1)
![{\displaystyle x\in \left[1,9\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c382323526a3226306c482a7e21229542942994b)
![{\displaystyle y\in \left[2,{\frac {10}{3}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/67f7bb544ddc2a11b3554f7b3e3045c62bfd10f9)
![{\displaystyle x\in \left[\pi ,{\frac {9}{4}}\pi \right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/98277867d64ec64dd50b50a028ca57eaf0b9006e)
![{\displaystyle \mu \in \left[\pi ,{\frac {9}{4}}\pi \right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/39312b6220ae96b3cbd7da6667e8205f57963de8)
